Maths Solutions Class 6 Chapter 7 Fractions Ganita Prakash

Figure it Out (Page 152)

Fill in the blanks with fractions.

1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh __1/3__ kg.

2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is __1/4__ kg.

3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ____ glass of sugarcane juice.

Solution : 

Amount of sugarcane consumed by four friends = 3 glasses .

Amount of sugarcane consumed by each friend = 34 glass Hence, each one drank 34 glass of sugarcane juice.

4. The bis fish weighs 12 kg. The small one weighs 14 kg. Together they weigh ____ kg.

Solution:

Given the weighs of big fish = 1/2 kg and the weighs of small fish = 1/4 kg

Total weight of both fish = 12 + 14 = 2+14kg

= 3/4  kg

Figure it Out (Page 155)

1. The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?

Solution:

(a) We get this piece by breaking the chikki into 12 equal parts. Hence it is 1/2

(b) We get this piece by breaking the chikki into 4 equal parts. Hence it is 1/4

(c) We get this piece by breaking the chikki into 8 equal parts. Hence it is 1/8

(d) We get this piece by breaking the chikki into 6 equal parts. Hence it is 1/6

(e) We get this piece by breaking the chikki into 8 equal parts. Hence it is 1/8

(f) We get this part by breaking the chikki into 6 equal pieces. Hence it is 1/6

(g) We get this part by breaking the chikki into 24 equal pieces. Hence it is 1/24

(h) We get this part by breaking the chikki into 24 equal pieces. Hence it is 1/24

Figure it Out (Page 158)

1. Continue this table of 12 for 2 more steps.

Solution:

2. Can you create a similar table for 14?

Solution: Yes, we can create a similar table for 1/4. Here

3. Make 13using a paper strip. Can you use this to also make 16 ?

Solution:

Fold the pa per strip along the length into 3 equal sections and each section represents 13.

Fold the pa per strip again in half breadthwise there by creating 6 equal sections and each section represents 16.

4. Draw a picture and write an addition statement as above to show:

(a) 5 times 14 of a roti

Solution:

(b) 9 times 14 of a roti

Solution:

5. Match each fractional unit with the correct picture :

Figure it Out (Page 160)

1. On a number line, draw lines of lengths 110,310,45.

Solution:

Step 1. Draw a line l. Mark a point O on it.

Step 2. Mark point A at a distance of 1 unit from O.
Step 3. Divide OA into 10 equal parts.
Here OP represents 110,

OQ represents 310, and

OR represents 810 = 45

2. Write five more fractions of your choice and mark them on the number line.

Solution:

Step 1. Let a number line OJ is divided into 10 equal parts.
Step 2. Now mark points A, B, C, D,…., J on it.
Step 3. Here

OF represents 610 = 35

OG represents 710

OB represents 210 = 15

OH represents 810 =45

OI represents 910

3. How many fractions lie between 0 and 1 ? Think, discuss with your classmates, and write your answer.

Solution:

There are an infinite number of fractions between 0 and 1.

Example: 35,45,710,12etc.

4. What is the length of the pink line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is 1/2. So the pink line is y units long. Write the fraction that gives the length of the black line in the box.

Solution:

Length of black line is 12

Length of black line is 12 + 12 +12

Fraction that gives length of black line = 32

5. Write the fraction that gives the lengths of the black lines in the respective boxes.

Solution:

Figure it Out  i (Page 162)

1. How many whole units are there in 72 ?

Solution: Here 72 = 7 times 12

= 1 + 1 + 1 + 12

= 3 + 12

= 312

Hence 3 whole units are there in 72.

2. How many whole units are there in 43 and in 73? 

Solution:

Here 43 = 4 times 13 = 13+13+13+13

=  1+1+13 + 13

= 1 + 13 = 113

Hence 1 whole unit are there in 43  and 73  = 7 times 13

Figure it Out  ii (Page 162)

Solution:

2. Can all fractions greater than 1 be written as such mixed numbers?
A mixed number /mixed fraction contains a whole number (called the whole part) and a fraction that is less than 1 (called the fractional part).

Solution:

Yes, all fractions greater than 1 can be written as mixed fractions/numbers.

3. Write the following fractions as mixed fractions (e.g.,  92 = 412)

Solution:

(a) 92

Figure it Out (Page 163)

1. Write the following mixed numbers as fractions:

Figure it Out (Page 165)

3. 46 = ___________ = ________ = ___________ = __________
(Write as many as you can)

Solution:
Here,

Figure it Out (Page 166)

1. Three rods are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts and multiplication facts.

Fraction of roti each child gets is _________________ .
Division fact:
Addition fact:
Multiplication fact:
Compare your picture and answers with your classmates!

Solution:

2. Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.

Solution:

As 2 rotis have to be shared equally by 4 children we divide each roti in 4 parts and give
(a) 1 part of each roti to each child as shown below:

Figure it Out (Page 168, 169)

1.Find the missing numbers:

(a) 5 glasses of juice shared equally among 4 friends is the same as _______ glasses of juice shared equally among 8 friends. So,54 = ?8.

(b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ______ bags. So, 43 = 12?.

(c) 7 rods divided among 5 children is the same as rods divided among children. So, 75 = _______

Solution:

(a) Here, the amount of juice each friend gets when 5 glasses are shared among 4 friends

= numberofglassesnumberoffriends=54

Now to determine how many glasses of juice would be needed to give each of the 8 friends the same amount = 8 × 5/4

= 10 glasses
So, 10 glasses of juice shared equally among 8 friends is the same as 5 glasses of juice shared equally among 4 friends.

Therefore 54 = 108

(Page 172)

1. Find equivalent fractions for the given pairs of fractions such that the fractional units are the same.

Figure it Out (Page 173)

1. Express the following fractions in lowest terms:

Figure it Out (Page 174)

1. Compare the following fractions and justify your answers:

(a) 83, 52

Solution:

(a) Given fractions are 83 and 52.

Here LCM of denominators 3 and 2 is 6 then multiplying and dividing 8/3 by 2 and 5/2 by 3 then

(b) 49, 37

Solution: Given fractions are 49 and37. Here LCM of denominators 9 and 7 is 63.

then multiplying and dividing 49 by 7 and 37 by 9 then

(c) 710, 914

Solution: Given fractions are 710 and 914. Here LCM of denominators 10 and 14 is 70.

then multiplying and dividing 710 by 7 and 914 by 5 then

2. Write following fractions ascending order.

(a) 710, 1115, 25

Solution: The given fractions are 710, 1115, 25 Let us find LCM of denominator 10, 15, 5

∴ LCM of 10, 15 and 5 = 2 × 3 × 5 = 30
Now let us make denominator of each fractions as LCM

Hence given fractions in ascending order are: 25, 710, 1115

(b) 1924, 56, 712

Solution:

The given fractions are 1924, 56, 712

Here LCM of 24, 6, 12 is 24.

On arranging in ascending Order, we get

1424, 1924, 2024

⇒ 712, 1924, 56

3. Write the following fractions descending order.

(a) 2516,78,134,1732,34,125,712,54

Solution:

(b) 34,125,712,54

Solution:

Figure it Out (Page 179)

1. Add the following fractions using Brahmagupta’s method:

(a) 27,57,67

Solution:

Here 27+57+67

(b) 34+13

Solution:

Here 34+13

Here LCM of denominators 4 and 3 is 12

∴ Equivalent fraction of 34 with denominators 12 is 912 and equivalent fraction of 13 with denominators 12 is 412

(c) 23+56

Solution:

Given  23+56

Now LCM of 3 and 6 is 6.
Expressing as equivalent fractions with denominators 6, we get

(d) 23+27

Solution:

Here 23+27

Now LCM of 3 and 7 is 21
Expressing as equivalent fractions with denominators 21, we get

(e) 34+13+15

Solution:

Here 34+13+15

Now LCM of 4, 3, 5 is 60.
Expressing as equivalent fractions with denominators 60, we get

(f) 23+45

Solution:

Here 23+45

Now LCM of 3 and 5 is 15
Expressing as equivalent fractions with denominators 15, we get

(g) 45+23

Solution:

Here 45+23

Now LCM of 5 and 3 is 15
Thus expressing as equivalent fractions with denominators 15, we get

(h) 33+58

Solution:

Given 33+58

Here LCM of 5 and 8 is 40
Expressing as equivalent fractions with denominators 40, we get

(i) 92+54

Solution:

Here 92+54

Now LCM of 2 and 4 is 4.
Expressing as equivalent fractions with denominators 4, we get

(j) 83+27

Solution:

Given 83+27

Here LCM of 3 and 7 is 21
Expressing as equivalent fractions with denominators 21, we get

(k) Same as (e) Part

(l) 23+45+37

Solution:

Here 23+45+37

Now LCM of 3, 5 and 7 is 105.
Expressing as equivalent fractions with denominators 105, we get

(m) 92+54+76

Solution:

Given 92+54+76

Here LCM of 2, 4, 6 is 12.
Now expressing as equivalent fractions with denominators 12, we get

2. Rahim mixes 23 litres of yellow paint with 34 litres of blue paint to make green paint. What is the volume of green paint he has made?

Solution:

Given quantity of yellow paint = 23 litres

and quality of blue paint = 34 litres

Volume of green paint made = 23 litres + 34 litres

Here LCM of 3 and 4 is 12
Now expressing as equivalent fractions with denominator 12, we get

3. Geeta bought 25 meter of lace and Shamim bought 34 meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?
Solution:
Given length of lace bought by Geeta = 25 meter

and length of lace bought by Shamim = 34 meter

Total length of lace = 25 + 34

LCM of 5 and 4 is 20
Now expressing as equivalent fractions with denominator 20, we get

Total lace required = Perimeter = 1 m
Hence total lace Geeta and Shamim have together is sufficient to cover the whole border.

Figure it Out (Page 181)

1. 58−38

Solution:

Given 58−38

As fractional unit is same i.e., 18 we shall simply subtract numerators keeping fractional unit as 518

Then 58−38 = 5−38

= 28=14(representing in simplest form)

2. 79−59

Solution:

Given 79−59

As fractional unit is same i.e., 519 we shall simply subtract numerators keeping fractional unit as 519
79−59
= 7−59 = \(\frac{2}{9}\)

1. Carry out the following subtractions using Brahmagupta’s method:

(a) 815−315

Solution:

(c) 56−49

 2. Subtract as indicated:

(a) 134from103

Solution:

Given y – y
Here, LCM of 3 and 4 is 12.
Fractional unit for both fractions should be  12

(b) 185from233

Solution:

Here, 233−185

Now, LCM of 3 and 5 is 15.

Fractional unit = 115 for both fractions
Hence

(c) 297from457

Solution:

Here fractional, 17 for both fractions

3. Solve the following problems:

(a) Java’s school is 710 km from her home. She takes an auto for 12 km from her home daily,

and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?

Solution:
Given distance between Jaya’s school and home is 710 km

and distance covered by Jaya in auto is 12 km.

∴ Distance Jaya covered by walking = 710−12 km

LCM of 10 and 2 is 10.